# Standing Wave Equation: Why does assuming a small slope $du/dx$ not make $d^2u/dx^2$ negligible as well?

Referencing the above image, just change the label for $$y$$-axis to $$u$$-axis.^

Following the derivation of the standing wave equation: https://www.youtube.com/watch?v=IAut5Y-Ns7g&t=1324s

So if slopes are small is an assumption: that means that du/dx~0. Wouldn't that make $$d2u/dx2$$ even more 0? Why would we remove $$du/dx$$ from the equation but not its second partial with respect to $$x$$?

Let's take an example signal: $$y=\cos(x)$$ about $$x=0$$.

$$\frac{dy}{dx}=-\sin(0)=0$$;

$$\frac{d^2y}{dx^2}=-\cos(0)=-1$$.

It is important to note that one is not simply the square of the other.

In the video, he is not assuming $$\frac{du}{dx}$$ is small. It has a definite nonzero value. He is taking the limit as $$\Delta x$$ gets small (goes to zero, or becomes an infinitesimal $$dx$$).

That expression

$$\lim_{\Delta x \rightarrow 0} \frac{1}{\Delta x} \left[\frac{du}{dx}_{(x+\Delta x)} -\frac{du}{dx}_{(x)} \right]$$

is the very definition of the derivative, so it becomes:

$$\frac{d}{dx} \left(\frac{du}{dx}\right)$$ or $$\frac{d^2u}{dx^2}$$

EDIT: Regarding your question on the "small slopes" assumption, small $$du/dx$$ does not necessarily mean small $$d^2u/dx^2$$. Taking a numerical example, you could imagine a sawtooth-like function whose slope $$du/dx$$ oscillates from $$-0.1$$ to $$0.1$$ over a $$\Delta x$$ of 0.01, which means its $$d^2u/dx^2$$ in that vicinity is 20.

In the vibrating string example, the "small slopes" assumption means that horizontal (x direction) motion of points on the string are negligible. Otherwise if you tried to pull the string up say to 45°, the string would have to change length or the endpoints would have to move, and the restoring force might not be linear, invalidating the simple wave model.

• Yes, I understand this. Apologies for not making my question more clear. What I'm confused about is that "slopes are small at all times" is assumption number 5 for the derivation (see 8:53 of his video). Doesn't that mean that du/dx=0 is a liberty being taken? Thus, the second partial of 0 is just 0? Jun 4 at 1:18

From my own experience in the past, there is a common beginner's mistake in interpreting notation. It makes a big difference if $$du/dx(x_0)=0$$ for a single location $$x_0$$ or $$du/dx(x)=0$$ for all $$x$$ in a continuous interval. Mathematicians are much more stringent in their notation, but in physics fine details are often swept under the rug (because the author thinks that they are self-evident).

In the former case, $$d^2u/dx^2(x_0)$$ can be anything, much like the constants $$a$$ and $$b$$ (as well as $$c$$, of course) in $$u(x)=ax^2+bx+c$$ are completely arbitrary. In the latter case, $$d^2u/dx^2(x)=0$$ for all $$x$$ in the interior of said interval.

The two entities, du/dx and d/dx(du/dx), don't have the same units. That means they cannot be compared. If one is 'small' in some sense, the other need not be. That second derivative, with the second power of 'x' in the denominator, can blow up just like any other divide-by-zero calculation.